# Line Integral over circle region

## Homework Statement

Evaluate ∫c (x + y) ds, where C is the circle centred at (1/2, 0) with radius 1/2.

## The Attempt at a Solution

parametrise
x=1/2cos(t)
y=1/2sin(t)
0≤t≤2π
ds=√dx2+dy2
=√(1/2)2-sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2(sin2(t)+cos2(t))
ds=1/2

∫ [(1/2)cos(t) + (1/2)sin(t)]*(1/2) dt, where 0≤t≤2π

I evaluated this integral and got 0. Is this because C is a simple closed curve?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Evaluate ∫c (x + y) ds, where C is the circle centred at (1/2, 0) with radius 1/2.

## The Attempt at a Solution

parametrise
x=1/2cos(t)
y=1/2sin(t)
0≤t≤2π
ds=√dx2+dy2
=√(1/2)2-sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2sin2(t)+(1/2)2cos2(t)
=√-(1)2(1/2)2(sin2(t)+cos2(t))
ds=1/2

∫ [(1/2)cos(t) + (1/2)sin(t)]*(1/2) dt, where 0≤t≤2π

I evaluated this integral and got 0. Is this because C is a simple closed curve?

I evaluated the (correct) integral and obtained an answer of ##\pi/n##, where ##n## is an integer whose exact value I will not divulge at this time. I assumed the circle was traversed in a counter-clockwise direction, which you did not mention in the problem statement.

The problem statement didn't mention the direction. So I just assume anti-clockwise.

Do you know where I have made my mistake? I have been on this problem for a while now and can't seem to find it. I've put the integral into a computation as well and got back 0 as the answer.

Ray Vickson
Homework Helper
Dearly Missed
The problem statement didn't mention the direction. So I just assume anti-clockwise.

Do you know where I have made my mistake? I have been on this problem for a while now and can't seem to find it. I've put the integral into a computation as well and got back 0 as the answer.

PF rules forbid me from telling you exactly where your mistake lies. All I can do is urge you to be more careful, and to read the question carefully.

Argh!

The circle is centred at (1/2,0) not at the origin. I'm not entirely sure how this changes anything though.

SteamKing
Staff Emeritus
Homework Helper
Argh!

The circle is centred at (1/2,0) not at the origin. I'm not entirely sure how this changes anything though.
Since C is a circle, perhaps a more natural expression for ds would be r ⋅ dθ

I found a source online that suggested if the circle is not at the origin then:
x = h + r cos(t)
y = k + r sin(t)
so
x = 1/2 + (1/2 cos(t))
y = 1/2 sin(t)
edit: FOUND THE INFO I WAS LOOKING FOR!

Circle with Center at Point (h,k)
http://www.regentsprep.org/regents/math/geometry/gcg6/LCirh3.gif
where the center is (h,k)

so

x= 1/2cos(t)-1/2
y= 1/2sin(t)

Ray Vickson
Homework Helper
Dearly Missed
I found a source online that suggested if the circle is not at the origin then:
x = h + r cos(t)
y = k + r sin(t)
so
x = 1/2 + (1/2 cos(t))
y = 1/2 sin(t)
edit: FOUND THE INFO I WAS LOOKING FOR!

Circle with Center at Point (h,k)
http://www.regentsprep.org/regents/math/geometry/gcg6/LCirh3.gif
where the center is (h,k)